Just so we're on the same page, here's what a right rotation about node $x$ will do:
Where $T, U, V$ are the subtrees of nodes $x$ and $y$. It's easy to see that if you start with a left-child chain consisting of nodes $x$ and $y$ as above (with $T,U,V$ empty), a right rotation about $x$ will yield the right-child chain in the right figure.
Now do the same thing, starting with a left-child chain of three nodes, $x,y,z$, reading from the top. If you do a right rotation about $y$ and then another about $x$, you'll have the following transformations:
and now if you recursively transform the $x,y$ left-child branch (since we know how to do that), you'll wind up with a right-child tree, $z,y,x$, reading from the top. It shouldn't be hard to see that this process will work for left-child chains of arbitrary length $n$, and will require $n(n-1)$ rotations, so no more than that number of right rotations will change a left-child tree to a right-child tree.
It's worth noting that, although the question didn't require it, this process will maintain the binary search tree property, giving the nodes in the reverse order of the original.